At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1920 m/s. The gas is

Options

A. `H_2`
B. `F_2`
C. `CI_2`
D. `O_2`

Answer

To solve the problem of identifying the diatomic gas based on its RMS speed, we will follow these steps:

### Step 1: Understand the formula for RMS speed
The root mean square (RMS) speed of gas molecules is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( V_{rms} \) = RMS speed of the gas molecules
- \( R \) = Universal gas constant (8.314 J/(mol·K))
- \( T \) = Absolute temperature in Kelvin
- \( M \) = Molar mass of the gas in kg/mol

### Step 2: Identify the known values
From the problem, we know:
- \( V_{rms} = 1920 \, \text{m/s} \)
- Room temperature \( T = 300 \, \text{K} \)
- \( R = 8.314 \, \text{J/(mol·K)} \)

### Step 3: Rearrange the formula to solve for \( M \)
We can rearrange the formula to isolate \( M \):
\[ M = \frac{3RT}{V_{rms}^2} \]

### Step 4: Substitute the known values into the equation
Now we substitute the known values into the rearranged formula:
\[ M = \frac{3 \times 8.314 \times 300}{(1920)^2} \]

### Step 5: Calculate the numerator
Calculating the numerator:
\[ 3 \times 8.314 \times 300 = 7482.6 \]

### Step 6: Calculate the denominator
Calculating the denominator:
\[ (1920)^2 = 3686400 \]

### Step 7: Calculate \( M \)
Now we can calculate \( M \):
\[ M = \frac{7482.6}{3686400} \approx 0.00203 \, \text{kg/mol} \]

### Step 8: Convert \( M \) to grams
To convert \( M \) to grams:
\[ M \approx 2.03 \, \text{g/mol} \]

### Step 9: Identify the gas based on molar mass
Now we compare the calculated molar mass with the options provided:
- \( H_2 \) = 2 g/mol
- \( F_2 \) = 38 g/mol
- \( Cl_2 \) = 71 g/mol
- \( O_2 \) = 32 g/mol

The calculated molar mass of approximately 2.03 g/mol corresponds to \( H_2 \).

### Conclusion
Thus, the gas is \( H_2 \).

---